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Quiz 3
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Find the value of λ such that the vectors \(\vec{a}\) = 2\(\hat{i}\) + λ\(\hat{j}\) + \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) are orthogonal
0%
0
0%
1
0%
\(\frac{3}{2}\)
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–\(\frac{5}{2}\)
Explanation
–\(\frac{5}{2}\)
The value of λ for which the vectors 3\(\hat{i}\) – 6\(\hat{j}\) + \(\hat{k}\) and 2\(\hat{i}\) – 4\(\hat{j}\) + λ\(\hat{k}\) are parallel is
0%
\(\frac{2}{3}\)
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\(\frac{3}{2}\)
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\(\frac{5}{2}\)
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–\(\frac{2}{5}\)
Explanation
\(\frac{2}{3}\)
The vectors from origin to the points A and B are \(\vec{a}\) = 2\(\hat{i}\) – 3\(\hat{j}\) +2\(\hat{k}\) and \(\vec{b}\) = 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) respectively, then the area of triangle OAB is
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340
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\(\sqrt{25}\)
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\(\sqrt{229}\)
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\(\frac{1}{2}\) \(\sqrt{229}\)
Explanation
\(\frac{1}{2}\) \(\sqrt{229}\)
For any vector \(\vec{a}\) the value of (\(\vec{a}\) × \(\vec{i}\))² + (\(\vec{a}\) × \(\hat{j}\))² + (\(\vec{a}\) × \(\hat{k}\))² is equal to
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\(\vec{a}\)²
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3\(\vec{a}\)²
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4\(\vec{a}\)²
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2\(\vec{a}\)²
Explanation
2\(\vec{a}\)²
If |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 2 and \(\vec{a}\).\(\vec{b}\) = 12, then the value of |\(\vec{a}\) × \(\vec{b}\)| is
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5
0%
10
0%
14
0%
16
Explanation
16
The vectors λ\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\), \(\hat{i}\) + λ\(\hat{j}\) – \(\hat{k}\) and 2\(\hat{i}\) – \(\hat{j}\) + λ\(\hat{k}\) are coplanar if
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λ = -2
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λ = 0
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λ = 1
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λ = -1
Explanation
λ = -2
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors such that \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = \(\vec{0}\), then the value of \(\vec{a}\).\(\vec{b}\) + \(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\)
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1
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3
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–\(\frac{3}{2}\)
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None of these
Explanation
–\(\frac{3}{2}\)
Projection vector of \(\vec{a}\) on \(\vec{b}\) is
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(\(\frac{\vec{a}.\vec{b}}{|\vec{b}|^2}\))\(\vec{b}\)
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\(\frac{\vec{a}.\vec{b}}{|\vec{b}|}\)
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\(\frac{\vec{a}.\vec{b}}{|\vec{a}|}\)
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(\(\frac{\vec{a}.\vec{b}}{|\vec{a}|^2}\))\(\hat{b}\)
Explanation
\(\frac{\vec{a}.\vec{b}}{|\vec{b}|}\)
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three vectors such that \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 5 and |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3, |\(\vec{c}\)| = 5, then the value of \(\vec{a}\).\(\vec{b}\) +\(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\) is
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0
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1
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-19
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38
Explanation
-19
If |\(\vec{a}\)| 4 and – 3 ≤ λ ≤ 2, then the range of |λ\(\vec{a}\)| is
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[0, 8]
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[-12, 8]
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[0, 12]
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[8, 12]
Explanation
[-12, 8]
The number of vectors of unit length perpendicular to the vectors \(\vec{a}\) = 2\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\) and \(\vec{b}\) = \(\hat{j}\) + \(\hat{k}\) is
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one
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two
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three
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infinite
Explanation
two
If (\(\frac{1}{2}\), \(\frac{1}{3}\), n) are the direction cosines of a line, then the value of n is
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\(\frac{\sqrt{23}}{6}\)
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\(\frac{23}{6}\)
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\(\frac{2}{3}\)
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–\(\frac{3}{2}\)
Explanation
\(\frac{\sqrt{23}}{6}\)
Find the magnitude of vector 3\(\hat{i}\) + 2\(\hat{j}\) + 12\(\hat{k}\)
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\(\sqrt{157}\)
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4\(\sqrt{11}\)
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\(\sqrt{213}\)
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9√3
Explanation
\(\sqrt{157}\)
Three points (2, -1, 3), (3, – 5, 1) and (-1, 11, 9) are
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Non-collinear
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Non-coplanar
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Collinear
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None of these
Explanation
Collinear
The vectors 3\(\hat{i}\) + 5\(\hat{j}\) + 2\(\hat{k}\), 2\(\hat{i}\) – 3\(\hat{j}\) – 5\(\hat{k}\) and 5\(\hat{i}\) + 2\(\hat{j}\) – 3\(\hat{k}\) form the sides of
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Isosceles triangle
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Right triangle
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Scalene triangle
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Equilateral triangle
Explanation
Isosceles triangle
The points with position vectors 60\(\hat{i}\) + 3\(\hat{j}\), 40\(\hat{i}\) – 8\(\hat{j}\) and a\(\hat{i}\) – 52\(\hat{j}\) are collinear if
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a = -40
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a = 40
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a = 20
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None of these
Explanation
a = -40
The ratio in which 2x + 3y + 5z = 1 divides the line joining the points (1, 0, -3) and (1, -5, 7) is
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5 : 3
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3 : 2
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2 : 1
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1 : 3
Explanation
5 : 3
If O is origin and C is the mid point of A (2, -1) and B (-4, 3) then the value of \(\bar{OC}\) is
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\(\hat{i}\) + \(\hat{j}\)
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\(\hat{i}\) – \(\hat{j}\)
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–\(\hat{i}\) + \(\hat{j}\)
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–\(\hat{i}\) – \(\hat{j}\)
Explanation
–\(\hat{i}\) + \(\hat{j}\)
If ABCDEF is regular hexagon, then \(\vec{AD}\) + \(\vec{EB}\) + \(\vec{FC}\) is equal
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0
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2\(\vec{AB}\)
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3\(\vec{AB}\)
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4\(\vec{AB}\)
Explanation
4\(\vec{AB}\)
If \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = 2\(\hat{i}\) – 4\(\hat{k}\), \(\vec{c}\) =\(\hat{i}\) + λ\(\hat{j}\) + 3\(\hat{j}\) are coplanar, then the value of λ is
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\(\frac{5}{2}\)
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\(\frac{3}{5}\)
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\(\frac{7}{3}\)
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–\(\frac{5}{3}\)
Explanation
–\(\frac{5}{3}\)
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